Again. Yes/No: do you claim the heated plate will remain at 150F after the second plate is added?
Repeat Latour's comment here:
However, the absorption rate of real bodies depends on whether the absorber T (radiating or not), is less than the intercepted radiation T, or not. If the receiver T [is less than] intercepted T, no [net] absorption occurs; if the receiver T [is less than] intercepted T the absorption rate may be as great as proportional to (T intercepted - T absorber), depending on the amounts reflected, transmitted or scattered.
(I have added [net] to indicate his argument is net heat transfer there, as Latour has explained many times elsewhere. I have also replaced "less than symbol" with [is less than] due to Slashdot's character restrictions.)
I repeat: you are completely (and wrongly) ignoring real-world conditions that apply to the experiment.
The experiment requires the passive plate to be some unspecified distance from the heat source. (This is a condition of the experiment; no contact is allowed since this is only about radiative transfer.) And no matter its configuration, it will radiate some of its absorbed energy outward to the chamber walls. This much we know.
The chamber walls C are actively cooled, although at a fixed power input. So it absorbs radiation from the internal space. We know at equilibrium T(c) will always be less than T at the source S: T(c) [is less than] T(s) because the wall is actively cooled. (We know for another reason too, but this is sufficient.)
We also know that the passive plate P will always be at a temperature less than that of the source, for the simple reason that no matter what its position, it does not absorb all the radiation from S. Or even if it did, as in the case of completely enclosing S, it would still re-radiate some of its absorbed energy out to the chamber walls C. Therefore as long as the conditions of the experiment are met, no matter what else varies such as relative mass, in this experiment T(p) will be lower than T(s). The amount is of no consequence, as long as it is non-zero (and it is).
Therefore, from S-B law, we can directly infer the following things:
T(c) [is less than] T(s) [net heat transfer is from source to walls, never the other way around]
T(p) [is less than] T(s) [net heat transfer is from source to plate, never the other way around]
We can also infer from the experimental conditions that T(c) [is less than] T(p), but that is irrelevant to the argument.
When equilibrium is achieved, these conditions still hold.
An elementary, obvious, and perfectly sound conclusion from this is that the source is not made hotter under any of these conditions. Even if plate P completely encloses source S, we know for at least two different reasons (greater radiative area, and the simple fact that it does radiate outward to the wall, which cools it) that T(p) will always be less than T(s), even at equilibrium.
Since the temperature of every other object is less than that of the heat source, there is no net heat flow TO the heat source, therefore the heat source does not become hotter. This is, and has been, the whole of Latour's argument, and it is valid. It is not crazy speculation by some nitwit, it is straightforward application of Stefan-Boltzmann law.
Q.E.D., indeed. If the above inequalities hold (and they do), Latour's conclusion is the only one that is mathematically valid.